In the canonical quantization of a massive Lagrangian, we obtain equations of motion via the Euler-Lagrange formalism. As I have read, these can become rather long but decouple into the Klein-Gordon (or Dirac) plus a set of lower-order equations. E.g., for the spin-1 Proca field: $$ (\Box + m^2) A^\mu - \partial^\mu (\partial^\nu A_\nu) = 0 \Rightarrow (\Box + m^2)A^\mu = 0,~ \partial^\mu A_\mu = 0. $$ As I see it, there is no obvious way of going back, i.e., deriving the coupled equations from the decoupled ones.

When constructing quantum fields of spin $s$ via irreducible representations of the Poincaré group (let's call it the Weinberg approach), we end up with completely symmetric tensors (or spinors, in the half-integer spin case) with $s$ Lorentz indices. As was shown in this answer, it follows that these fields fulfill a number of differential equations, namely: $$ (\Box + m^2) \Phi_{\mu_1\cdots\mu_s} = 0 \text{ or } (i\not\partial + m) \Phi_{\mu_1\cdots\mu_s} = 0 ~,~~ \partial^{\mu_1} \Phi_{\mu_1\cdots\mu_s} = 0 \text{ or } \gamma^{\mu_1} \Phi_{\mu_1\cdots\mu_s} = 0~,~~ \eta^{\mu_i\mu_j} \Phi_{\mu_1\cdots\mu_s} = 0, $$
depending on whether it is a tensor or tensorial spinor. So here, we directly obtain the decoupled equations of motion.

My question now is: Are the "coupled" equation of motion an artifact of the Lagrange formalism or is there a way to derive them from the Weinberg approach?

I am curious about this because time-ordered products of the fields do have a special relation to the coupled equations of motion. They are the respective greens function. E.g., for the Proca field: $$ \Big(\eta^{\mu\nu} (\Box + m^2) - \partial^\mu \partial^\nu \Big) \langle T A_\nu(x) A_\rho(x') \rangle = -i \eta^\mu_\rho \delta(x-x'). $$ This does not necessarily follow from the decoupled equations, even though we have $D(\partial) \langle T A_\nu(x) A_\rho(x') \rangle \propto \delta(x-x')$ for $D = (\Box + m^2), \partial^\mu, ... $.

I am happy about any insight on how these properties are related!

[Edit:] With some help, I was able to narrow down what I am looking for. Let me try to (hopefully) clarify.

It is always possible to combine the "decoupled" equations such that the resulting equations are equivalent to the ones we started with. The equations of motion resulting from the Lagrangian are special because they allow to describe a deformation with an interaction Lagrangian $\mathcal{L}$ via a simple Dyson-series(?) (due to renormalization, this holds only when coupling linealy to an external potential, more general interactions will deviate), e.g. for spin-0 or spin-1: $$ (\Box + m^2)\phi_{int} = - \Big( \frac{\partial \mathcal{L}}{\partial \phi}\Big)_{int},~ \Big(\eta^\mu_\nu (\Box + m^2) - \partial^\mu \partial_\nu \Big) A^\nu = - \Big( \frac{\partial \mathcal{L}}{\partial A_\mu}\Big)_{int}. $$ The $_{int}$ here denotes a perturbative series in the propagators, here the time-ordered products (this corresponds to the equation for the propagator of $A^\mu$ I wrote above).

So my question would rather be:

Is there a generalization of this formalism to higher-spin fields? Are there still equations of motion and time-ordered products that act as their Green's functions?

Thanks again for any help!

  • $\begingroup$ For free massive fields the situation becomes more complicated for higher-spin. For a Lagrangian realisation of a spin $s$ and mass $m$ rep of the Poincare group you need auxiliary fields of spin $s-2,s-3,\dots,0$ in addition to the spin $s$ parent field (see the famous work by Singh and Hagen, Fierz and Pauli for spin 2, or the Stuckelberg approach by Zinoviev). Each field has its own ('coupled') EoM and you need to eliminate all auxilliary fields on-shell. $\endgroup$
    – SigmaAlpha
    Dec 15, 2020 at 14:36
  • $\begingroup$ For (half-)integer spin massless fields you have the (Fang-)Fronsdal Lagrangian, which only requires spin ($s$-1 and) $s-2$ auxilliary fields. In this case you have the higher-spin gauge symmetry which shortens the required tower of auxiliary fields and simplifies the analysis. $\endgroup$
    – SigmaAlpha
    Dec 15, 2020 at 14:38

2 Answers 2


Several comments.

  1. This is not really a well-defined question. Say I have field equations $\mathcal A\phi=\mathcal B\phi=0$ for some differential operators $\mathcal A,\mathcal B$. This system is obviously equivalent to $(\mathcal A+\mathcal B)\phi=(\mathcal A-\mathcal B)\phi=0$. It is also equivalent to $(\mathcal A\phi)^2+(\mathcal B\phi)^2=0$. Which of these presentations is "coupled" and which is "decoupled"? There is no answer: they are all equivalent. If I write down an arbitrary differential equation, how do I determine whether it is "coupled" or "decoupled"? Until you give a formal definition of what it means to be "(de)coupled", I don't think your question really makes sense. Right now it is merely based on aesthetics. One presentation "looks better" than the other, but this is not a precise notion.

  2. Most PDEs do not admit a Lagrangian description, and those that do, we in general don't know its Lagrangian to begin with. Figuring out a suitable Lagrangian given a set of differential equations is a rather non-trivial task. And this Lagrangian is never unique. So in general there is no "canonical approach". The equations of motion are well-defined; the Lagrangian is most often not, and when it is it is not unique. In this sense, the "Weinberg" approach is the most general one.

  3. In the case of arbitrary spin fields, the equations of motion are known as the Fronsdal equations. I think these were first written down in 1978. A valid Lagrangian was obtained a year later, 1979. This Lagrangian was not even guaranteed to exist to begin with. In fact, in the interacting case we still don't know of a Lagrangian description, and I believe most people suspect it doesn't exist (at least in the traditional sense; there are some proposals in terms of "master fields" similar in spirit to String field theory).

  4. In an arbitrary theory, the propagator always satisfies the equations of motion up to contact terms. As the "coupled" and "decoupled" forms are equivalent, the propagator will satisfy both, so both approaches are equally as natural. Again, I don't think there is a well-defined distinction between what you call "coupled" and "decoupled"; they are literally the same equations, written in different form.

  5. In the case of Proca you are missing a term, $\eta^{\mu\nu}\partial^2-\partial^\mu\partial^\nu+\eta^{\mu\nu}m^2$. The mass term is essential: by taking a derivative you get $m^2(\partial\cdot \Delta)\sim 0$, so for $m\neq0$ you get the standard transversality condition. In the massless case this condition disappears. But anyway, I am not sure what point you are trying to make regarding the Proca field. The two presentations of the equations of motion are equivalent so they lead to the same propagator, even if one approach is perhaps less transparent in terms of what degrees of freedom are dynamical. The on-shell (Weinberg) approach was designed precisely to make these d.o.f. manifest. (By contrast, the off-shell (Lagrangian) approach was designed to make locality manifest, and also because it is straightforward to quantise).

  • $\begingroup$ Regarding 1. They are not equivalent, it only goes in one direction. Consider for example $\phi(x)$ a differentiable function, $\mathcal{A} = \partial_x $ and $\mathcal{B} =1$. Then, $\mathcal{B} \phi(x) = \phi(x) = 0$, but $(\mathcal{A} \phi(x))^2 + (\mathcal{B} \phi(x))^2 = 0$ has a non-trivial solution. I was asking whether one can express several differential equations as a single equivalent, covariant expression (as it would result from the Lagrange formalism). Your comment regarding the Fronsdal equations solves that. It is not possible. Auxiliary fields an several equations are needed. $\endgroup$
    – Cream
    Dec 15, 2020 at 15:01
  • $\begingroup$ Regarding 4. and 5. "the propagator always satisfies the equations of motion up to contact terms" Yes, but at least for the Proca field (also the massive scalar and Dirac field) I see a distinction. The "coupled" equation of motion exactly inverts the propagator exactly inverts the propagator as a Green's function would, one applied to the other gives a Dirac delta. The decoupled e.o.m. are satisfied up to contact terms, but can contain linear combinations of derivatives of $\delta$. I am asking if this is a special feature for some form of the e.o.m. for higher spin fields. $\endgroup$
    – Cream
    Dec 15, 2020 at 15:11
  • $\begingroup$ @Cream Re.1: note that a sum of squares is zero iff each square itself is zero. If you have a collection of equations, you can always gather them into a single equation, for example by squaring and adding. Please have a look at this somewhat famous lecture of Feynman, section 25–6. $\endgroup$ Dec 15, 2020 at 15:26
  • $\begingroup$ The concept of "number of equations" is not well-defined: you can always rewrite any system in different equivalent forms, each having a different number of equations. You can reduce it to a single equation (the great “law of nature", in Feynman words), but also the other way around: you can recast a single equation as an infinite system of equations. In the higher-spin community this is a rather useful thing to do, and is known as the unfolded formulation. It is interesting to read about it, if you want. $\endgroup$ Dec 15, 2020 at 15:27
  • $\begingroup$ Re.Re.1: This holds for real numbers. The above example would admit the solution $\phi(x) = e^{ix}$, or am I missing something? But you are right that it is always possible to combine e.o.m. by adding the absolute value squared without loosing information. I realized that I am actually looking for something more specific; I edited the question to (hopefully) clarify. This unfolded formulation sounds interesting in any case. I will take a look. $\endgroup$
    – Cream
    Dec 16, 2020 at 11:01

Let's discuss for simplicity only free massive fields. Usually fields can interpolate for different representations of the Poincaré group. Equations of motion are equivalent to projectors on states you want to describe (including the on-shell condition). For instance

  • Scalar field. It can interpolate only for spin $0$ states, therefore, equation of motion is \begin{equation} (\Box+m^2)\phi=0. \end{equation}

  • Vector field. It can interpolate for spin $0$ and spin $1$ states. The corresponding projectors (you can easily convince yourself) are \begin{equation} \Pi^L_{\mu \nu} = \frac{\partial_\mu \partial_\nu}{\Box}, ~~\text{and} ~~ \Pi^T_{\mu \nu} = \eta _{\mu \nu}-\Pi^L_{\mu \nu}=\eta _{\mu \nu}-\frac{\partial_\mu \partial_\nu}{\Box} \end{equation}

    1. To get spin $1$ we thus have to impose \begin{equation} \Pi^L_{\mu \nu} A^\nu = 0 ~~ \text{and} ~~ (\Box+m^2)\Pi^T_{\mu \nu} A^\nu = 0. \end{equation} Obviously, due to properties of projectors (orthogonality), a linear combination of the two constraints will be equivalent to the two equations. In other words, we can write \begin{equation} \left [ \alpha \Pi^L_{\mu \nu} + (\Box+m^2)\Pi^T_{\mu \nu} \right ] A^\nu = 0. \end{equation} Is it possible now to obtain this equation from a local Lagrangian? Yes, using the definition of projectors we get for $\alpha = m^2$ \begin{equation} m^2\frac{\partial_\mu \partial_\nu}{\Box} + (\Box+m^2) \left ( \eta _{\mu \nu} - \frac{\partial_\mu \partial_\nu}{\Box}\right )= \eta_{\mu \nu}(\Box+m^2)-\partial_\mu \partial_\nu, \end{equation} leading to \begin{equation} 0 = (\Box+m^2)A_\mu-\partial_\mu (\partial A) =m^2A_\mu - \partial^\nu F_{\mu \nu}, \end{equation} which is clearly derivable from the Lagrangian \begin{equation} \mathcal L_1 = -\frac{1}{4}F^2+\frac{1}{2}m^2 A^2. \end{equation}
    2. To get spin $0$ we need to impose \begin{equation} \Pi^T_{\mu \nu} A^\nu = 0 ~~ \text{and} ~~ (\Box+m^2)\Pi^L_{\mu \nu} A^\nu = 0. \end{equation} Similarly, considering a linear combination \begin{equation} m^2\Pi^T_{\mu \nu} + (\Box+m^2)\Pi^L_{\mu \nu} = m^2\eta_{\mu \nu}+\partial_\mu \partial_\nu, \end{equation} resulting in \begin{equation} 0 = m^2 A_\mu + \partial _\mu (\partial A), \end{equation} which can be obtained from \begin{equation} \mathcal L_0 = \frac{1}{2}(\partial A)^2-\frac{1}{2}m^2 A^2. \end{equation} In all these cases getting the Lagrangian is simple, you just multiply eom by the field itself (it does not spoil the orthogonality).
  • You can do the same for say Dirac field which interpolates only for spin $1/2$ states. In this case the Dirac equation is just a constraint (and there is no need to impose $p^2=m^2$).

Similar procedures can be done for any spin.

  • $\begingroup$ Okay, so longitudinal modes have to vanish, transverse modes have to propagate according to KG. Questions: 1. Are there only two projections for any spin or more? I can imagine there are several long. and transv. directions? 2. As there is no Lagrangian without auxiliary fields for spin > 3/2, how does this procedure change? Are the projections of the field not zero but instead equal the auxiliary field (which vanishes on-shell)? 3. Does this relate to the edit of my question? Meaning: If we introduce an interaction, how are the long. & transv. projections of the field deformed? $\endgroup$
    – Cream
    Dec 17, 2020 at 14:21
  • 1
    $\begingroup$ @Cream "so longitudinal modes have to vanish, transverse modes have to propagate according to KG". That depends on what representation you want to single out. 1. No there are as many projectors as there are representations that a certain field can interpolate for. 2. There are Lagrangians for spin $>3/2$, say for spin $2$ (Fierz-Pauli). 3. I have nothing to say about higher spins with interaction, except that there exists Vasiliev construction $\endgroup$
    – nwolijin
    Dec 17, 2020 at 19:59

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